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\textbf{Question}

In this question YOU MAY ASSUME
\begin{description}
\item{(i)}
that small changes $df$ in the function $f(S,t)$ are related to small
changes in $S$ and $t$ by Taylor's theorem so that
$$df = f_SdS +f_tdt + \frac{1}{2} f_{SS}dS^2 + f_{St}dSdt +
\frac{1}{2} f_{tt} dt^2+ \cdots$$

\item{(ii)}
that $S$ follows the lognormal random walk
$$\frac{dS}{S} = rdt + \sigma dX$$
where $r$ and $\sigma$ are constants and $X$ is a random variable,

\item{(iii)}
that $dX^2 \to dt$ as $dt \to 0$.
\end{description}

\begin{description}
%Question 3a
\item{(a)}
Derive It\^{o}'s lemma in the form
$$df=\sigma S f_S dX + \left ( f_t + rSf_S + \frac{1}{2} \sigma^2 S^2
f_{SS} \right ) dt$$
and comment briefly on whether or not your derivation is rigorous.

%Question 3b
\item{(b)}
Denote the fair value of an option by $V(S,t)$. By constructing a
portfolio $\Pi = V - \Delta S$ where $\Delta$ is to be determined,
show that $V$ satisfies the Black-Scholes equation
$$V_t + \frac{1}{2} \sigma^2 S^2 V_{SS} + r SV_S - rV =0.$$

%Question 3c
\item{(c)}
A PERPETUAL option is one whose value does not depends upon time. Find
the most general solution for the value of a perpetual option and show
that the value of a perpetual Put is given by
$$V=AS^{-2r/\sigma^2}$$
where $A$ is a constant that depends on the specific details of the
option.
\end{description}

\newpage 
\textbf{Answer}

\begin{description}
%Question 3a
\item{(a)}
We have, by Taylor's theorem:-
$$df=f_SdS+f_tdt+\frac{1}{2}f_{SS}dS^2+ f_{St}dSdt+
\frac{1}{2}f_{tt}dt^2+\cdots$$
and 
\begin{eqnarray*} dS &= &S\mu dt+ S\sigma dX\\
\Rightarrow \ dS^2 & = & S^2\mu^2dt^2+2S^2\mu\sigma dtdX+S^2\sigma^2dX^2
\end{eqnarray*}

But now as $dt\to 0$, $dX^2\to dt$
\begin{eqnarray*}
\Rightarrow dS^2 & = & S^2\sigma^2dt +2\sigma \mu S^2(dt)^{3/2}+
S^2\mu^2dt^2\\
& = & S^2\sigma^2dt+O(dt^{3/2})\\
\Rightarrow df & = & f_S[S\mu dt+ S\sigma dX]+f_tdt+
\frac{1}{2}(f_{SS}\sigma^2 S^2 dt\\
& & + O(dt^{3/2}))+ O(dt^{3/2})
\end{eqnarray*}
and so, to leading order,
\begin{eqnarray*}
df & = & f_S[S\mu dt+ S\sigma dX]+f_tdt+\frac{1}{2}\sigma^2 S^2 f_{SS}
dt\\
& = & S\sigma F_S dX +(F_t+\mu Sf_S+ \frac{1}{2}\sigma^2 S^2f_{SS})dt
\ \ - \ \ \rm{ITO's\ lemma}
\end{eqnarray*}

The derivation is not very rigorous at all - it started from Taylor's
theorem which is valid for smooth functions - and S follows a random
walk!


%Question 3b
\item{(b)} 
Now consider the portfolio $\Pi=V-S\Delta$.

We have
\begin{eqnarray*}
d\Pi & = & dV-\Delta dS = S\sigma V_S dX+ \left ( V_t+\mu SV_S+
\frac{1}{2} \sigma^2 S^2 V_{SS} \right ) dt\\
& & - \Delta (\mu Dst+\sigma
SdX)\\
d\Pi & = & (S\sigma V_S-\Delta S \sigma)dX\\
& &  + \left ( V_t+\mu SV_s +
\frac{1}{2} \sigma^2 S^2 V_{SS}- \Delta S\mu \right ) dt
\end{eqnarray*}

All the randomness in $\Pi$ may thus be eliminated by choosing
$\Delta=V_S$, in which case we find that
\begin{eqnarray*}
d\Pi & = & \left ( V_t + \mu SV_S + \frac{1}{2}\sigma^2 S^2 V_{SS}
-S\mu V_S \right ) dt\\
& = & \left ( V_t +\frac{1}{2} \sigma^2 S^2 V_{SS} \right ) st.
\end{eqnarray*}

We now appeal to arbitrage: presumably the option must be neither more
nor less valuable than a risk free investment, otherwise one or the
other would never be used. So the above must be equal to the return in
time $dt$ of an amount $\Pi$ invested in a risk free portfolio. Thus
\begin{eqnarray*}
r\Pi dt & = & \left ( V_t + \frac{1}{2} \sigma^2 S^2 V_{SS} \right )
dt\\
r(V-S\Delta) & = & V_t +\frac{1}{2} \sigma^2 S^2 V_{SS}\\
r(V-SV_S) & = & V_t + \frac{1}{2}\sigma^2 S^2 V_{SS} \ \ \Rightarrow \
\rm{Black-Scholes}\\
\end{eqnarray*}
$$V_t +\frac{1}{2}\sigma^2 S^2 V_{SS} +rSV_S -rV = 0.$$ 


%Question 3c
\item{(c)}
For a perpetual option we have $V_t=0$ $\Rightarrow\ V=V(S)$ only.

$\Rightarrow$ Black-Scholes becomes
$$\frac{1}{2} \sigma^2 S^2 V_{SS} +rSV_S -rV =0$$
This is Euler's equation, so for solution try $V=S^k$

\begin{eqnarray*}
\frac{1}{2} \sigma^2 S^2 k(k-1)S^{k-2} +rSkS^{k-1} -rS^k & = & 0\\
S^k \left ( \frac{1}{2} \sigma^2 k(k-1) +rk -r \right ) & = & 0
\end{eqnarray*}

So for a solution we need $\frac{1}{2} \sigma^2 k(k-1) +(k-1)r=0$.

$\begin{array}{llr}
\Rightarrow & \rm{either} \ \ & k=1\\
& \rm{or} & \frac{1}{2} \sigma^2 k +r=0\\
\Rightarrow & & k=-2r/\sigma^2
\end{array}$

Thus the most general solution for a perpetual option is
$$V=AS+BS^{-2r/\sigma^2}$$
where $A$ and $B$ are arbitrary constants.

Now consider an American (or European) Put which is perpetual! Clearly
as $S\to\infty$ the option becomes more and more worthless, since the
chance of exercising it becomes less and less. $\Rightarrow A=0$

$\Rightarrow$ for some $\overline{A}$
$$V+\overline{A}S^{-2r/\sigma^2}$$

\end{description}

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